Quantum Electrodynamics : Ãœbungsblatt 0

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Quantum Electrodynamics : Ãœbungsblatt 0

Quantum Electrodynamics Prof. John Schliemann, Sebastian Waeber SS 19 Sheet 12 Task 30: The regularized polarization function On the last sheet we used Pauli-Villars regularization to write the regularized polari- reg 2 zation tensor iΠµν as the gauge invariant combination (q gµν − qµqν) times e2 Λ2 i ln + Π(¯ q 2); (1) 12π2 m2 where m is the mass of the electron, Λ is the momentum scale and Π¯ is a finite integral over the parameter β, introduced on the last sheet. More specifically Π¯ there turned out to be proportional to Z 1 dββ(1 − β) ln[1 − β(1 − β)q 2=m2]: (2) 0 In this task we are going to solve this integral. a) Perform a partial integration to get rid of the logarithm and introduce the new integration variable v = 2β − 1 to write the integrand as a rational function of v 2. b) Consider the integral Z 1 1 dv 2 : (3) 0 v − c In the regions c ≤ 0 and in the region 1 ≤ c you can solve this integral easily. For 0 < c < 1 we remember a trick, when treating Feynman-propagators: In order to treat poles on the integration curve we introduce a tiny imaginary part c ! c + i, which we send to 0 in the end. We can split the integration curve into a term comming from the residuum theorem and a principle value integral. Derive the solution to (3) for the different regions of c. c) To treat the integral deduced in a), we consider Z 1 v n In = dv 2 : (4) 0 v − c − i Find a recursiv relation between In and In−2 and use this to write down an explicit expression for Π¯ for the three cases considered in b) now with respect to q 2. Task 31: The Uehling-Potential reg The higher order correction to the polarization tensor Πµν gives rise to a quantum field theoretic correction to the classical Coulomb-potential, which is known as the Uehling potential. For this we consider the source jµ = −Zeδ(~x)δµ0 (5) 1 of a stationary pointlike charge Z · e. For the first order (in α) corrected photon 1 propagator DF,µν(q): 1 reg DF,µν(q) = DF,µν(q) + DF,µa (q)Πab DF;bν(q) (6) we have the relation Z d 4q A0 (x) = e−iqx D 1 (q)j ν(q); (7) µ (2π)4 F,µν ν with the elecromagnetic current j (q) and DF,µν(q) is the free photon propagator. Derive the Uehling potential using your results from the last sheet and the previous task. Give your result depending on the renormalized charge. 2.

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